乘积大于或等于 0 的最长子数组的长度

给定一个由N 个整数组成的数组arr[] ，任务是找到乘积大于或等于 0 的最长子数组的长度。
例子：

Input: arr[] = {-1, 1, 1, -2, 3, 2, -1 } 
Output: 6 
Explanation: 
The longest subarray with product ≥ 0 = {1, 1, -2, 3, 2, -1} and {-1, 1, 1, -2, 3, 2}. 
Length of each = 6.

Input: arr[] = {-1, -2, -3, -4} 
Output: 4 
Explanation: 
The longest subarray with product ≥ 0 = {-1, -2, -3, -4}. 
Length = 4. 

方法：

1. 检查给定数组中所有元素的乘积是否大于或等于零。
2. 如果是，则乘积大于或等于零的最长子数组的长度就是数组的长度。
3. 如果上述陈述不成立，则数组包含奇数个负元素。在这种情况下，要找到最长的子数组，请执行以下操作：
   • 对于数组中出现的每个负元素，当前元素左边和右边的子数组给出大于或等于 0 的乘积。因此所需的最长子数组的长度为：

L = max(L, max(i, N - i - 1))

• 为数组中找到的每个负元素不断更新子数组的长度。
• L的值是乘积大于等于 0 的最长子数组的长度。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
 
#include 
using namespace std;
 
// Function that count the length
// of longest subarray with product
// greater than or equals to zero
int maxLength(int arr[], int N)
{
    int product = 1, len = 0;
 
    for (int i = 0; i < N; i++) {
        product *= arr[i];
    }
 
    // If product is greater than
    // zero, return array size
    if (product >= 0) {
        return N;
    }
 
    // Traverse the array and if
    // any negative element found
    // then update the length of
    // longest subarray with the
    // length of left and right subarray
    for (int i = 0; i < N; i++) {
        if (arr[i] < 0) {
            len = max(len,
                      max(N - i - 1, i));
        }
    }
 
    return len;
}
 
// Driver Code
int main()
{
    int arr[] = { -1, 1, 1, -2, 3, 2, -1 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << maxLength(arr, N) << endl;
 
    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;
 
public class GFG{
// Function that count the length
// of longest subarray with product
// greater than or equals to zero
    static int maxLength(int arr[], int N)
    {
        int product = 1, len = 0;
     
        for (int i = 0; i < N; i++) {
            product *= arr[i];
        }
     
        // If product is greater than
        // zero, return array size
        if (product >= 0) {
            return N;
        }
     
        // Traverse the array and if
        // any negative element found
        // then update the length of
        // longest subarray with the
        // length of left and right subarray
        for (int i = 0; i < N; i++) {
            if (arr[i] < 0) {
                len = Math.max(len, Math.max(N - i - 1, i));
            }
        }
     
        return len;
    }
     
    // Driver Code
    public static void main(String args[])
    {
        int arr[] = { -1, 1, 1, -2, 3, 2, -1 };
        int N = arr.length;
        System.out.println(maxLength(arr, N));
     
    }
}
 
// This code is contributed by AbhiThakur

Python3

# Python3 implementation of the above approach
 
# Function that count the Length
# of longest subarray with product
# greater than or equals to zero
def maxLength(arr, N):
    product = 1
    Len = 0
 
    for i in arr:
        product *= i
 
    # If product is greater than
    # zero, return array size
    if (product >= 0):
        return N
 
    # Traverse the array and if
    # any negative element found
    # then update the Length of
    # longest subarray with the
    # Length of left and right subarray
    for i in range(N):
        if (arr[i] < 0):
            Len = max(Len,max(N - i - 1, i))
 
    return Len
 
# Driver Code
if __name__ == '__main__':
    arr = [-1, 1, 1, -2, 3, 2, -1]
    N = len(arr)
 
    print(maxLength(arr, N))
 
# This code is contributed by mohit kumar 29

C#

// C# implementation of the above approach
using System;
 
class GFG{
// Function that count the length
// of longest subarray with product
// greater than or equals to zero
    static int maxLength(int []arr, int N)
    {
        int product = 1, len = 0;
     
        for (int i = 0; i < N; i++) {
            product *= arr[i];
        }
     
        // If product is greater than
        // zero, return array size
        if (product >= 0) {
            return N;
        }
     
        // Traverse the array and if
        // any negative element found
        // then update the length of
        // longest subarray with the
        // length of left and right subarray
        for (int i = 0; i < N; i++) {
            if (arr[i] < 0) {
                len = Math.Max(len, Math.Max(N - i - 1, i));
            }
        }
     
        return len;
    }
     
    // Driver Code
    public static void Main()
    {
        int []arr = { -1, 1, 1, -2, 3, 2, -1 };
        int N = arr.Length;
        Console.WriteLine(maxLength(arr, N));
     
    }
}
 
// This code is contributed by abhaysingh290895

Javascript

 
// Javascript implementation of the above approach
 
// Function that count the length
// of longest subarray with product
// greater than or equals to zero
function maxLength(arr, N)
{
    var product = 1, len = 0;
 
    for (var i = 0; i < N; i++) {
        product *= arr[i];
    }
 
    // If product is greater than
    // zero, return array size
    if (product >= 0) {
        return N;
    }
 
    // Traverse the array and if
    // any negative element found
    // then update the length of
    // longest subarray with the
    // length of left and right subarray
    for (var i = 0; i < N; i++) {
        if (arr[i] < 0) {
            len = Math.max(len,
                      Math.max(N - i - 1, i));
        }
    }
 
    return len;
}
 
// Driver Code
var arr = [ -1, 1, 1, -2, 3, 2, -1 ];
var N = arr.length;
document.write(maxLength(arr, N));
 
// This code is contributed by rutvik_56.

输出：

6

时间复杂度： O(N) ，其中 N 是数组的长度。
辅助空间： O(1)